An octagon which has side lengths 3, 3, 11, 11, 15, 15, 15 and 15 is inscribed
in a circle. Find the area of the octagon.
since the central angles must be a complete 2pi. The angles of 3,11,15,15 must sum to pi.
Half of the angles must look like arcsin(3/(2x))
Asking Wolfram for the radius x that solves
arcsin(3/(2x))+arcsin(11/(2x))+2arcsin(15/(2x))=pi/2
gives
x=5sqrt(17/2) [This agrees with Charlie. Also confirmed graphically]
x^2=425/2
The apothem, h, to each side, s, is then of the form
h^2=x^2(s/2)^2
s=3, h=29/2
s=11, h=27/2
s=15, h=25/2
Finally sum 1/2*s*h*2 (to go all the way around)
3*29/2+11*27/2+2*15*25/2 = 567
Note to Charlie's solution: The circle will have area pi*425/2 = 667.6
and this octagon is roughly the same as in inscribed regular hexagon which would have an area of about 552.1. The answer should be in this ballpark.
Note on my solution: I attempted to solve for x analytically using the sum of arcsines formula, but the algebra got to be too much.

Posted by Jer
on 20240613 18:35:45 